Topics in Advanced Mathematics: Topology
Homework 3a – Solutions
Wojciech Komornicki
18 Febrauary 2014
2.2 3 Let X be a topological space, and let A ⊆ X. A point x in X is called limit point of A if each neighbourhood
b and
b Prove that A = A ∪ A,
of x contains points of A other than x. The set of limit points of A is written A.
b ⊆ A. Give examples of non-empty subsets A if R such that
that A is closed ⇔ A
b = ∅.
(i) A
Example A = Z.
b 6= ∅ and A
b ⊆ A.
(ii) A
b = A.
If A = [0, 1] then A
b
(iii) A is a proper subset of A.
b = [0, 1].
Example If A =]0, 1[ then A
Example
b 6= ∅ but A ∩ A
b = ∅.
(iv) A
Example
b = {0}.
Let A = {1/n : n a positive integer} . Then A
b
Proof: First, we prove that A = A ∪ A,
Let a ∈ A. Then for any neighbourhood N of a, N ∩ A 6= ∅. If a 6∈ A, a 6∈ N ∩ A. Hence, if a 6∈ A,
b
a ∈ A.
b
Therefore, A ⊆ A ∪ A,
b ⊆ A. Hence A ∪ A
b⊆A
By definition, A ⊆ A and A
b
Therefore A = A ∪ A,
Now suppose that A is closed.
b and so A
b ⊆ A.
Then A = A. Hence A = A = A ∪ A
b⊆A
Suppose that A
b = A and so A is closed.
Then A = A ∪ A
2.2 5 Let I = [0, 1]. Define an order relation ≤ on I2 = I × I by
(x, y) ≤ (x′ , y ′ ) ⇔ y < y ′ or (y = y ′ and x ≤ x′ ).
The television topology on I 2 is the order topology (see problem 2.1 2 on page 21) with respect to ≤ (the name
is due to E.C. Zeemann). Let A be the set of points (2−1 , 1 − n−1 ) for positive integral n. Prove that in the
television topology on I2
A = A ∪ {(0, 1)}.
Proof: Note that if a, b, c ∈ I with a < b, then
](a, c), (b, c)[= {(x, y) : a < x < b, y = c} .
Also if a, b, c, d ∈ I with a < b and c < d then
](a, c), (b, d)[= {(x, c) : x > a} ∪ {(x, y) : 0 ≤ x ≤ 1, c < y < d} ∪ {(x, d) : x < b}
If (0, 1) ∈](a, c), (b, d)[ where a, b, c, d ∈ I, then (a, c) < (0, 1) and (0, 1) < (b, d). Hence c < 1, d = 1 and
0 < b. Hence A∩](a, c), (b, d)[ so (0, 1) ∈ A.
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Moreover, if (x, y) ∈ A\I2 then there exist a, b, c, d ∈ I such that (a, c) < (x, y) < (b, d) but ](a, c), (b, d)[∩A =
∅.
Hence
A = A ∪ {(0, 1)}.
2.2 8 For any subset A of a topological space X, define ExtA (the exterior of A), BdA (the boundary of A) and
FrA (the frontier of A) as follows
ExtA = Int(X\A)
Bd A = A\IntA
Fr A = Bd A ∪ Bd(X\A).
Prove that the following relations hold:
(i) A = IntA ∪ Fr A = A ∪ Fr A = A ∪ Bd(X\A).
Proof:
A = X\Int(X\A)
= A ∪ ((X\A)\Int(X\A))
= A ∪ Bd(X\A)
= Int A ∪ (A\Int A) ∪ Bd(X\A)
= Int A ∪ Fr A
⊆ A ∪ Fr A
= A ∪ Bd A ∪ Bd(X\A)
⊆ A ∪ Bd(X\A)
⊆A
Hence A = IntA ∪ Fr A = A ∪ Fr A = A ∪ Bd(X\A).
(ii) Int(Bd A) = ∅.
Proof:
Let a ∈ Int(Bd A). Then a ∈ Int A →←
Hence Int(Bd A) = ∅.
(iii) Bd(IntA) = ∅.
Proof:
Bd(IntA) = Int A\Int(Int A) = Int A\Int A = ∅.
(iv) Bd(Bd A) = Bd A.
Proof:
Bd(Bd A) = Bd A\Int(Bd A)
= Bd A\∅
= Bd A\
Hence Bd(Bd A) = Bd A.
(v) Fr A = A ∩ (X\A).
Proof:
A ∩ (X\A) = X\Int(X\A) ∩ X\Int A = X\ (Int(X\A) ∪ Int A) = Fr A
(vi) Fr A is closed. If A is closed then Bd A = Fr A.
Proof:
Fr A = Bd A ∩ Bd(X\A) so Fr A is closed.
Suppose that A is closed. Then Int(X\A) = X\A. Hence Bd(X\A) = ∅.
Therefore Fr A = Bd A ∪ Bd(X\A) = Bd A.
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(vii) Fr(Fr(Fr A)) = Fr(Fr A) ⊂ Fr A.
Proof:
Since Fr A is closed
Fr(Fr(Fr A)) = Bd(Bd(Fr A)) = Bd(Fr A) = Fr(Fr A)
Also, for any S, Bd S ⊆ S. Hence Fr(Fr A) ⊂ Fr A.
(viii) Fr A = ∅ ⇔ A is both open and closed.
Proof:
Suppose Fr A = ∅
Then Bd A = ∅ and Bd(X\A) = ∅.
Since Bd A = ∅, A is open.
Since Bd(X\A) = ∅, X\A is open and so A is closed.
Now suppose that A is both open and closed.
Then Bd, A = ∅ and Bd(x\A) = ∅ and so Fr A = ∅.
(ix) Bd A = A ∩ (X\A).
Proof:
By 2.2.5
A ∩ (X\A) = A ∩ (X\Int A) = A\Int A = Bd A
(x) Bd A is closed ⇔ A is the union of a closed and an open set.
Proof:
Suppose Bd A is closed.
Then
A = Bd A ∪ Int A
and so A is the union of a closed and an open set.
Now suppose that A = U ∪ V where U is open and V is closed.
Then, since U ⊂ Int A,
Bd A = A ∩ (X\A) = (U ∪ V ) ∩ (X\A) = V ∩ (X\A).
Hence Bd A is closed.
(xi) Ext(Ext A) = Int A.
Proof:
Let a ∈ Ext(Ext A)
Note that Ext A = Int(X\A) = X\A.
Hence
Ext(Ext A) = Int(X\Ext A) = Int(X\(X\A)) = Int A
(xii) Ext Ext Ext Ext A = Ext Ext A.
Proof:
Clearly Ext Ext Ext Ext A ⊆ Ext Ext A.
Let a ∈ Ext Ext A = Int A. Then there exists an open set U such that a ∈ U ⊆ A.
Hence a ∈ U ⊆ Int A ⊆ Int A.
Therefore a ∈ Int Int A. Hence Ext Ext A ⊆ Ext Ext Ext Ext A and so Ext Ext Ext Ext A = Ext Ext A.
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